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We want to find the integral of a function at an arbitrary location x from the origin. Thus, where I(x=0) is the value of the integral for all times less than 0. (Essenti
If y 1 (t) and y 2 (t) are two solutions to y′′ + p (t ) y′ + q (t ) y = 0 So the Wronskian of the two solutions is, W(y 1 ,y 2 )(t) = =
How to find total no. of unordered pairs of disjoint subsets of a finite set? Solution) Suppose A and B are two such disjoint subsets of the set S. Then every element can go into
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the median of a continuous frequency distribution is 21.if each observation is increased by 5. find the new median
∫1/sin2x dx = ∫cosec2x dx = 1/2 log[cosec2x - cot2x] + c = 1/2 log[tan x] + c Detailed derivation of ∫cosec x dx = ∫cosec x(cosec x - cot x)/(cosec x - cot x) dx = ∫(cosec 2 x
Given f (x) = - x 2 + 6 x -11 determine each of the following. (a) f ( 2) (b) f ( -10) (c) f (t ) Solution (a) f ( 2) = - ( 2) 2 + 6(2) -11 = -3 (
Let D(subscript12) = ({x,y : x^2 = e ; y^6 = e ; xy =(y^-1) x}) a) Which of the following subsets are subgroups of D(subscript12) ? Justify your answer. i) {x,y,xy,y^2,y^3,e}
arrange these numbers in ascending order. -5 -7 1 2 15 0 - 25
Prove the subsequent Boolean expression: (x∨y) ∧ (x∨~y) ∧ (~x∨z) = x∧z Ans: In the following expression, LHS is equal to: (x∨y)∧(x∨ ~y)∧(~x ∨ z) = [x∧(x∨ ~y)] ∨ [y∧(x∨
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